# Integration by substitution (and esp. Weierstrass' substitution)

#### dodo

The stupid question of the day.

Is it fair to say that$$\frac{du}{dx} = \frac 1 {dx/du}$$since this comes (I think) from the chain rule,$$\frac{dx}{du} \frac{du}{dx} = \frac{dx}{dx} = 1$$

Which means that, when integrating by substitution, I can choose to do either of$$\int f(u) du = \int f(x) \frac {du}{dx} dx = \int \frac {f(x)} {dx/du} dx$$depending on which derivative I happen to have at hand.

(Just checking; you can't be too careful when treating differentials as it they were fractions, which they aren't.)

The matter came out while studying Weierstrass' substitution, where\begin{align*} \int \frac 1 {\sin x} dx &= \int \frac {\frac{1+t^2}{2t}} {\frac{1+t^2}{2}} = \int \frac 1 t dt &\mbox{with }t=\tan \frac x 2 \end{align*}because $\frac {dt}{dx}$ is easier to figure out than $\frac {dx}{dt}$.

Any more detail (particularly from the viewpoint of analysis) is welcome. Right now I feel like I'm just shuffling symbols without really knowing what am I doing.

#### HallsofIvy

Homework Helper
As long as u(x) is invertible (i.e. as long as du/dx is not 0), yes, $dx/du= 1/(du/dx)$.

Your second statement is also correct but you have to be careful with it. Unless du/dx is a constant we cannot just shift du/dx inside or outside the integral. But we can use one or the other of those forms, whichever is appropriate.

#### dodo

Hey, Halls, thanks.
Unless du/dx is a constant we cannot just shift du/dx inside or outside the integral.
Yes, I absolutely agree! The intention was, as you say next, choosing one form or the other, depending on which derivative I find easier to get.